This may be resolved into a radial and a latitudinal component; i.e.,

F_r = −Fcos(φ) = − (dm)(rcos²(φ))Ω²
F_φ = Fsin(φ) = (dm)(rcos(φ)sin(φ))Ω²
 

The mass in an infinitesimal volume dV is ρdV where ρ is the fluid density. An infinitesimal volume between r and r+dr, φ and φ+dφ, and θ and θ+dθ is r²cos(φ)drdφθ. This can be represented as an area times a thickness in three ways:

dV = (rcos(φ)drdθ)(rdφ) = dA(rdφ)
dV = (r²cos(φ)dφθ)dr = dBdr
dV = (rdφdr)(rcos(φ)dθ) = dC((rcos(φ)dθ)
 

The Latitudinal Distribution of Pressure

The force on the infinitesimal element due the pressure p on the face at φ+dφ less the force on face at φ is

dA(∂p/∂φ)dφ
 

When this force is added to the latitudinal component of the centrifugal force on dV the result must be zero; i.e.,

−(∂p/dφ)dφdA + (ρdV)((rcos(φ)sin(φ))Ω²) = 0
which reduces to
∂p/∂φ = ρr²cos(φ)sin(φ)Ω² = ρr²(sin(2φ)/2)Ω²
 

Integrating this last equation with respect to φ from 0 to Φ with r constant gives:

p(Φ,r) - p(0,r) = ρr²Ω²[(cos(2Φ)-1)/4]
 

Therefore the equilibrium latitudinal pressure distribution is given by

p(Φ,r) = p(0,r) − ρr²Ω²[(1 − cos(2Φ))/4]
 

The Radial Distribution of Pressure

The area of the top and bottom surfaces of the infinitesimal volume are approximately equal to dB = r²cos(φ)dφθ. The infinitesimal volume is then dV=dBdr.

The difference in pressure on the top and bottom surfaces plus the gravitational pull on the infinitesimal volume plus the centrifugal force must equal zero; i.e.,

−(∂p/∂r)drdB −(ρdV)[g + rcos²(φ)Ω²] = 0
which reduces to
(∂p/∂r) = −ρ[g + rcos²(φ)Ω²]

When this last equation is integrated with respect to r from R₀ to R the result is:

p(φ,R) − p(φ,R₀) = −ρg(R−R₀) −ρ(cos²(φ)Ω²)(R²−R₀²)/2

This result presumed the gravitational attraction is constant with altitude, R−R₀. More generally the result would be the pressure as a function of altitude with a correction for the effect of the centrifugal force. The correction term depends upon the cosine squared of the latitude.